B. Case of Fake Numbers( Codeforces Round #310 (Div. 2) 简单题)

B. Case of Fake Numbers

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Andrewid the Android is a galaxy-famous detective. He is now investigating a case of frauds who make fake copies of the famous Stolp's gears, puzzles that are as famous as the Rubik's cube once was.

Its most important components are a button and a line of n similar gears. Each gear has n teeth containing all numbers from 0 to n - 1 in the counter-clockwise order. When you push a button, the first gear rotates clockwise, then the second gear rotates counter-clockwise, the the third gear rotates clockwise an so on.

Besides, each gear has exactly one active tooth. When a gear turns, a new active tooth is the one following after the current active tooth according to the direction of the rotation. For example, if n = 5, and the active tooth is the one containing number 0, then clockwise rotation makes the tooth with number 1 active, or the counter-clockwise rotating makes the tooth number 4 active.

Andrewid remembers that the real puzzle has the following property: you can push the button multiple times in such a way that in the end the numbers on the active teeth of the gears from first to last form sequence 0, 1, 2, ..., n - 1. Write a program that determines whether the given puzzle is real or fake.

Input

The first line contains integer n (1 ≤ n ≤ 1000) — the number of gears.

The second line contains n digits a1, a2, ..., an (0 ≤ ai ≤ n - 1) — the sequence of active teeth: the active tooth of the i-th gear contains number ai.

Output

In a single line print "Yes" (without the quotes), if the given Stolp's gears puzzle is real, and "No" (without the quotes) otherwise.

Sample test(s)

input

3
1 0 0

output

Yes

input

5
4 2 1 4 3

output

Yes

input

4
0 2 3 1

output

No

Note

In the first sample test when you push the button for the first time, the sequence of active teeth will be 2 2 1, when you push it for the second time, you get 0 1 2.

   题意：有n个转盘，在位置上分别为1-n，每一个转盘上在一圈分别写着0 -- n-1，

如今有一个button，每一次按这个button。每一个转盘都会发生一些变化，详细的变化是：

1.转盘在奇数位的。转盘上面的数字+1（也就是顺时针旋转）。

2.转盘在偶数位的，转盘上面的数字-1（也就是逆时针旋转）。

问能不能在摁下n次button后，使得第i个转盘上面的数字就是i-1。

    思路：推断一下第一个转盘为0时。其它的转盘是不是满足第i个转盘的数字为i-1。不满足直接输出No。

点击打开链接

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<stack>

using namespace std;

int n;
int a[1005];

int main() {
    while(scanf("%d",&n)!=EOF) {
        for(int i=1; i<=n; i++) {
            scanf("%d",&a[i]);
        }
        while(a[1] != 0) {
            for(int i=1; i<=n; i++) {
                if(i%2 == 1) {
                    a[i] += 1;
                    a[i] = a[i]%n;
                } else {
                    a[i] -= 1;
                    if(a[i]<0){
                        a[i] = n + a[i];
                    }
                }
            }
        }
        int flag = 0;
        for(int i=1;i<=n;i++){
            if(a[i] != i-1)
            {
                flag = 1;
                break;
            }
        }
        if(flag == 1){
            printf("No
");
        }
        else{
            printf("Yes
");
        }
    }
    return 0;
}